It's well known that there is an analogy for the Euler line in a tetrahedron, but is there also an analogy for the nagel line of a tetrahedron? I can't seem to find any decent literature talking about it - specifically, is it true that the incenter, centroid, and Spieker center of a tetrahedron lie on a line?
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$\begingroup$ By Spieker center you mean the barycentre of the surface of a tetrahedron? $\endgroup$– Fedor PetrovCommented Jun 8 at 10:56
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For a simplex $T\subset \mathbb{R}^d$, we may define its Spieker center $S$ as the barycenter of the surface of $T$. Then $S$, the centroid $G$ and the incentre $I$ lie on a line in this order, and $IG:GS=d:1$. Indeed, if $A_i$ are the vertices of $T$, and $B_i$ the barycentres of corresponding faces, then we may group the total mass $s_i$ of $i$-th face at $B_i$, thus $S$ is the barycenter of the system of material points $(B_i,s_i)$. Also, $I$ is the barycenter of the system of material points $(A_i,s_i)$. The simplex $T_0$ with vertices $B_i$ is homothetic to $T$ with respect to $G$ with coefficient $-1/d$. Thus, this homothety maps $I$ to $S$, so the claim.
answered Jun 8 at 11:07
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$\begingroup$ Would it be alright to ask for any known references or literature? $\endgroup$ Commented Jun 8 at 19:22
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1$\begingroup$ It is alright of course to ask, but I do not know:) $\endgroup$ Commented Jun 8 at 20:19